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Magazine Chapter 6: Problem 35
Find \(f_{x x}, f_{x y}, f_{y x},\) and \(f_{y y}\). $$f(x, y)=e^{2 x y}$$
Short Answer
Expert verified
\(f_{xx}=4y^2 e^{2xy}\), \(f_{xy}=f_{yx}=2x e^{2xy} + 4xy^2 e^{2xy}\), \(f_{yy}=4x^2 e^{2xy}\).
Step by step solution
01
Find the First Partial Derivative with respect to x
To find the derivative of \(f(x, y) = e^{2xy}\) with respect to \(x\), we apply the chain rule. The derivative of \(e^{u}\) is \(e^u\cdot u'\), where \(u=2xy\) and \(u'\) is 2y with respect to \(x\). Therefore, \(f_x = 2y \, e^{2xy}\).
02
Find \(f_{xx}\)
To find the second partial derivative with respect to \(x\), denoted as \(f_{xx}\), we differentiate \(f_x = 2y \, e^{2xy}\) with respect to \(x\) again. Apply the product rule: \((uv)' = u'v + uv'\), where \(u=2y\) and \(v=e^{2xy}\). This gives \(f_{xx} = 4y^2 \, e^{2xy}\).
03
Find the First Partial Derivative with respect to y
To find the derivative of \(f(x, y) = e^{2xy}\) with respect to \(y\), use the chain rule again. The derivative of \(u=2xy\) with respect to \(y\) is 2x. Therefore, \(f_y = 2x \, e^{2xy}\).
04
Find \(f_{xy}\)
To find the mixed partial derivative \(f_{xy}\), differentiate \(f_x = 2y \, e^{2xy}\) with respect to \(y\). Using the product rule, where \(u=2y\) and \(v=e^{2xy}\), we get \(f_{xy} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy}\).
05
Find \(f_{yx}\)
Since mixed partial derivatives are equal if the function is continuous and differentiable, \(f_{xy} = f_{yx}\). Hence, \(f_{yx} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy}\).
06
Find \(f_{yy}\)
To find the second partial derivative with respect to \(y\), denoted as \(f_{yy}\), differentiate \(f_y = 2x \, e^{2xy}\) with respect to \(y\) again. The product rule gives \(f_{yy} = 4x^2 \, e^{2xy}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
When dealing with partial derivatives, the chain rule is a valuable tool. It helps us in differentiating composite functions. In our exercise, we applied the chain rule to function \( f(x, y) = e^{2xy} \). Here's how it works:
- The chain rule states that to differentiate a composite function, we need to multiply the derivative of the outer function by the derivative of the inner function.
- For example, in the partial derivative of \( e^{2xy} \) with respect to \( x \), the inner function \( u = 2xy \) has a derivative \( u' = 2y \). Thus, employing the chain rule gives us \( f_x = 2y \cdot e^{2xy} \).
Product Rule
The product rule comes into play when differentiating products of functions. When you have a function like \( f(x, y) = 2y \, e^{2xy} \), this rule helps us find derivatives accurately.
- The product rule formula is expressed as \( (uv)' = u'v + uv' \).
- In Step 2, when finding \( f_{xx} \), we considered \( u = 2y \) and \( v = e^{2xy} \). Differentiating \( u \) gives us \( u' = 0 \), since \( 2y \) is constant regarding \( x \), while the derivative of \( v \) using the chain rule results in \( 2y \, e^{2xy} \).
- This results in \( f_{xx} = 4y^2 \, e^{2xy} \).
Mixed Partial Derivatives
Mixed partial derivatives show how a function changes when different variables are varied sequentially. In our example, \( f_{xy} \) and \( f_{yx} \) reflect the changes with respect to \( x \) and \( y \), respectively.
- First, differentiate \( f_x \) with respect to \( y \), using the product rule, as explored here, results in \( f_{xy} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy} \).
- In the same essence, \( f_{yx} \) is obtained by differentiating \( f_y \) with respect to \( x \), which also results in \( f_{yx} = 2x \, e^{2xy} + 4xy^2 \, e^{2xy} \).
- The equality \( f_{xy} = f_{yx} \) holds, assuming function continuity, due to Clairaut's theorem. This symmetry is critical in multivariable calculus, allowing for simplification in many cases.
